Question: Let $h(x)=\csc(x)$. Find $h'\left(\dfrac{\pi}{4}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $\sqrt{2}$ (Choice C) C $1$ (Choice D) D $-\sqrt{2}$
Explanation: Let's first find $h'(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{4}$. Recall that the derivative of $\csc(x)$ is $-\dfrac{\cos(x)}{\sin^2(x)}$, or $-\csc(x)\cot(x)$. [Is there a way to know this without memorizing?] So $h'(x)=-\dfrac{\cos(x)}{\sin^2(x)}$. Now let's plug $x={\dfrac{\pi}{4}}$ into $h'$ : $\begin{aligned} &\phantom{=}h'\left({\dfrac{\pi}{4}}\right) \\\\ &=-\dfrac{\cos\left({\dfrac{\pi}{4}}\right)}{\sin^2\left({\dfrac{\pi}{4}}\right)} \\\\ &=-\dfrac{\dfrac{\sqrt{2}}{2}}{\left(\dfrac{\sqrt{2}}{2}\right)^2} \\\\ &=-{\dfrac{\sqrt{2}}{2}}\cdot{\dfrac{2}{1}} \\\\ &=-\sqrt{2} \end{aligned}$ In conclusion, $h'\left(\dfrac{\pi}{4}\right)=-\sqrt{2}$.